unique-paths/unique-paths-ii

题目一：unique-paths

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

How many possible unique paths are there?

代码如下：

class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m,vector<int>(n,0));
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(i==0 || j==0)
dp[i][j]=1;
else
dp[i][j]=dp[i-1][j]+dp[i][j-1];
}
return dp[m-1][n-1];
}
};

题目二：unique-paths-ii

Follow up for “Unique Paths”:

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as1and0respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[

[0,0,0],

[0,1,0],

[0,0,0]

]

The total number of unique paths is2.

Note: m and n will be at most 100.

代码如下：

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m=obstacleGrid.size();
int n=obstacleGrid[0].size();
vector<vector<int>> dp(m,vector<int>(n,0));
for(int i=0;i<m;i++)
if(obstacleGrid[i][0]==1)
break;
else
dp[i][0]=1;
for(int j=0;j<n;j++)
if(obstacleGrid[0][j]==1)
break;
else
dp[0][j]=1;
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
{
if(obstacleGrid[i][j]==1)
dp[i][j]=0;
else
dp[i][j]=dp[i-1][j]+dp[i][j-1];
}
return dp[m-1][n-1];
}
};